Search results

Page title matches

  • ==Problem== ...rom <math>0</math> to <math>3</math> gives <math>1+11^2+11^{3}\cdot 10+11^{4}\cdot 10\cdot 9</math>. Notice that <math>10+9\cdot11\cdot10=10+990=1000</m
    4 KB (551 words) - 10:32, 5 February 2022
  • ==Problem== ...can do some cases on the number of 5's rolled (note that there are <math>6^4 = 1296</math> outcomes).
    20 KB (3,220 words) - 03:24, 14 August 2024

Page text matches

  • ...ake. Sometimes, the administrator may ask other people to sign up to write problems for the contest. * Look at past [[AMC]]/[[AHSME]] tests to get a feel for what kind of problems you should write and what difficulty level they should be.
    51 KB (6,175 words) - 21:41, 27 November 2024
  • ==Problem== This problem is essentially asking how many ways there are to choose <math>2</math> dist
    5 KB (831 words) - 18:47, 29 January 2025
  • ...n McNugget Theorem''' (or '''Postage Stamp Problem''' or '''Frobenius Coin Problem''') states that for any two [[relatively prime]] [[positive integer]]s <mat ...t Theorem has also been called the Frobenius Coin Problem or the Frobenius Problem, after German mathematician Ferdinand Frobenius inquired about the largest
    17 KB (2,823 words) - 23:06, 15 November 2024
  • ...eorems concerning [[polygon]]s, and is helpful in solving complex geometry problems involving lengths. In essence, it involves using a local [[coordinate syst ...school students made it popular. The technique greatly simplifies certain problems.
    5 KB (812 words) - 15:43, 1 March 2025
  • ...h> objects in <math>3</math> bins. The number of ways to do such is <math>{4+3-1 \choose 3-1} = {6 \choose 2} = 15</math>. ...ach urn, then there would be <math>{n \choose k}</math> possibilities; the problem is that you can repeat urns, so this does not work.<math>n</math> and then
    5 KB (795 words) - 17:39, 31 December 2024
  • ...oks like this when he is mad: https://cdn.artofproblemsolving.com/images/7/4/b/74b21fec95225e1621c71ec8f17f343c48a726e0.jpg GMAAS permits to post this p ...into anything. Using that fact, you can use the Games theorem to solve any problem.
    69 KB (11,805 words) - 20:49, 18 December 2019
  • ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2018 AIME II Problems|Entire Test]]
    1 KB (133 words) - 18:13, 18 March 2020
  • ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2019 AIME I Problems|Entire Test]]
    1 KB (133 words) - 17:41, 29 March 2019
  • ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2019 AIME II Problems|Entire Test]]
    1 KB (133 words) - 15:43, 22 March 2019
  • ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2020 AIME I Problems|Entire Test]]
    1 KB (133 words) - 18:11, 18 March 2020
  • {{AIME Problems|year=2018|n=II}} ==Problem 1==
    9 KB (1,385 words) - 00:26, 21 January 2024
  • Here are the problems from the 2019 AMC 10C, a mock contest created by the AoPS user fidgetboss_4000. ==Problem 1==
    12 KB (1,917 words) - 12:14, 29 November 2021
  • ==Problem== D = A+1/4*(B-A);
    7 KB (1,053 words) - 14:58, 14 January 2024
  • ==Problem== defaultpen(linewidth(0.4)+fontsize(10));
    7 KB (1,129 words) - 16:27, 6 January 2025
  • ==Problem== ...ing <math>2019</math>, and <math>f\left(\tfrac{1+\sqrt{3}i}{2}\right)=2015+2019\sqrt{3}i</math>. Find the remainder when <math>f(1)</math> is divided by <m
    4 KB (706 words) - 22:18, 28 December 2023
  • ==Problem== ...pretty. Let <math>S</math> be the sum of positive integers less than <math>2019</math> that are <math>20</math>-pretty. Find <math>\tfrac{S}{20}</math>.
    3 KB (474 words) - 01:38, 22 December 2024
  • ==Problem== <math>\frac{360}{7}^{\circ}</math> does not work because <math>\tan\left(4 \cdot \frac{360}{7}^{\circ}\right)</math> is positive.
    8 KB (1,172 words) - 18:21, 8 August 2024
  • ==Problem== label("$\omega_1$",(-4.5,1));
    14 KB (2,229 words) - 14:57, 27 December 2024
  • ==Problem== <cmath>r^2= 1^2 + \left(\sqrt{2}+1\right)^2= 4+2\sqrt{2}.</cmath>
    7 KB (1,051 words) - 20:45, 27 January 2024
  • ==Problem== ...used to obtain <math>96</math>, as we can use at most <math>\frac{96}{24}=4</math> stamps without going over. The potential sets are <math>(24, 25), (3
    9 KB (1,543 words) - 19:48, 3 January 2025

View (previous 20 | next 20) (20 | 50 | 100 | 250 | 500)